Let me start with 25 = 1 (mod 8) first, it's easier.
This is shorthand for saying that the remainder when we divide 25 by 8 is 1. Another example would be 5 = 2 (mod 3). When you divide 5 by 3 you get a remainder of 2.
Now, when you divide 7 by 8 you get a remainder of, well, 7. So 7 = 7 (mod 8). Where do I get the -1 from? Well, we're playing around with something called an "equivalence relation." In actuality I shouldn't be using the "=" symbol, it should have 3 bars across, not 2. But since LaTeX is still down I'm using it for shorthand.
Anyway, this -1 business. We can show that a remainder of 7 (mod 8) is equivalent to a remainder -1 (mod 8). Really it isn't that hard. Let's look at 25 = 1 (mod 8) again. When we do 25/8 we pick the largest integer "k" such that 8k < 25, then we get the remainder by r = 25 - 8k. In this case k = 3. BUT what if we took k = 4 and found r = 25 - 8k? Now r = -7. So 25 = 1 (mod 8) = -7 (mod 8).
This is all a way of leading you into the more general definition of this "modular" number system. When we say 25 = r (mod 8) what we are saying is that r = 25 - 8k, where k is some integer. This defines a set that r can be equal to. Any member of this set is said to be "equivalent a modulo 8."
Typically we choose positive numbers for r, but to make things easier we occasionally also pick negative numbers, as I did to solve the problem in this thread.
This really isn't the best introduction to the subject. (I'm not a bad writer, but neither am I a good one! ) But it isn't really that hard, it just takes a little getting used to. You might find this link to be helpful. The first section is probably all that you need, though if you know the algebra the rest of it is decently informative.
Hope it helps!
Exactly what course does this question come from?
If you aren't familiar with modular arithmetic, it's not a fair question.
What is the remainder when is divided by 8?
For reference, let
Any multiple of 1000 is divisible by 8.
Hence, we are concerned with only the last three digits of
We will indicate " ends in " with: .
 We find that for
 We find that:
. . . . . .